I don’t think the Elimination approach gives P(Heads|Awake) = 1⁄3 or P(Monday|Awake) = 2⁄3 in the Single Awakening problem. In that problem, there are 6 possibilities:
Elimination argument is the logic of “There are four equiprobable mutually exclusive outcomes but only 3 of them are compatible with an observation, so we update to 3 equiprobable outcomes” attempted to be applied to the Sleeping Beauty problem. This logic itself is correct when there actually are four equiprobable mutually exclusive outcomes. Like in a situation where two coins were tossed and you know that the resulting outcome is not Heads Heads.
So the argument tries to justify, why the same situation is true for the Sleeping Beauty problem. It does so based on these facts:
P(Heads) = P(Tails) - because the coin is fair
P(Monday)=P(Tuesday) - because the experement lasts for two days
Days happen independently of the outcomes of the coin tosses
The thing is, it’s not actually enough. And to show it I apply this reasoning to another prblem where all the same conditions are satisfied, yet the answer is definetely not 1⁄3.
What you’ve done is not an application of Elimination argument to Single-Awakening problem. You are trying to apply a version of Updating model to Single-Awakening problem. And just as Updating model when attempted to apply to Sleeping Beauty is actually solving Observer problem, you are solving Observer-Single-Awakening problem:
You were hired to work as an observer for one day in a laboratory which is conducting Single-Awakening experiment. You don’t know whether it’s the first day of the experiment or the second, but you can see whether the Beauty is awake or not.
You seem to be doing it correctly. No update happens because P(Heads|Awake)=P(Tails|Awake).
I don’t think the Elimination approach gives P(Heads|Awake) = 1⁄3 or P(Monday|Awake) = 2⁄3 in the Single Awakening problem. In that problem, there are 6 possibilities:
P(Heads&Monday) = 0.25
P(Heads&Tuesday) = 0.25
P(Tails&Monday&Woken) = 0.125
P(Tails&Monday&Sleeping) = 0.125
P(Tails&Tuesday&Woken) = 0.125
P(Tails&Tuesday&Sleeping) = 0.125
Therefore:
P(Heads|Awake)
= P(Heads&Monday) / (P(Heads&Monday) + P(Tails&Monday&Woken) + P(Tails&Tuesday&Woken))
= 0.5
And:
P(Monday|Awake)
= (P(Heads&Monday) + P(Tails&Monday&Woken)) / (P(Heads&Monday) + P(Tails&Monday&Woken) + P(Tails&Tuesday&Woken))
= 0.75
Elimination argument is the logic of “There are four equiprobable mutually exclusive outcomes but only 3 of them are compatible with an observation, so we update to 3 equiprobable outcomes” attempted to be applied to the Sleeping Beauty problem. This logic itself is correct when there actually are four equiprobable mutually exclusive outcomes. Like in a situation where two coins were tossed and you know that the resulting outcome is not Heads Heads.
So the argument tries to justify, why the same situation is true for the Sleeping Beauty problem. It does so based on these facts:
P(Heads) = P(Tails) - because the coin is fair
P(Monday)=P(Tuesday) - because the experement lasts for two days
Days happen independently of the outcomes of the coin tosses
The thing is, it’s not actually enough. And to show it I apply this reasoning to another prblem where all the same conditions are satisfied, yet the answer is definetely not 1⁄3.
What you’ve done is not an application of Elimination argument to Single-Awakening problem. You are trying to apply a version of Updating model to Single-Awakening problem. And just as Updating model when attempted to apply to Sleeping Beauty is actually solving Observer problem, you are solving Observer-Single-Awakening problem:
You seem to be doing it correctly. No update happens because P(Heads|Awake)=P(Tails|Awake).